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From mechanics to thermodynamics – analysis ofselected examples

J. Guemeza,1, M. Fiolhaisb,2

a Departamento de Fısica Aplicada

Universidad de Cantabria

E-39005 Santander, Spain

b Departamento de Fısica and Centro de Fısica Computacional

Universidade de Coimbra

P-3004-516 Coimbra, Portugal

Abstract

We present and discuss a selected set of problems of classical mechanics and thermodynamics.The discussion is based on the use of the impulse-momentum equation simultaneously with thecentre-of-mass (pseudo-work) equation or with the first law of thermodynamics, depending onthe nature of the problem. Thermodynamical aspects of classical mechanics, namely problemsinvolving non-conservative forces or variation of mechanical energy are discussed, in differentreference frames, in connection with the use of one or the other energy equation, and with thecompliance of the Principle of Relativity.

1 Introduction

Relations between mechanics and thermodynamics have been several times pointed out and peda-

gogically discussed in the literature. In particular we have studied in detail systems whose physics

description requires both mechanical and thermodynamical approaches such as the cartesian diver

[1] and the drinking bird [2]. Also many toys, whose physics was revisited in a recent publication

[3], require a mechanical-thermodynamical analysis in order to fully understand their behaviours.

However, in this work we want to emphasize the use of two “energy equations”, namely the pure

mechanical one, relating the variation of the centre-of-mass kinetic energy with the pseudo-work

performed by the resultant external force, and the other one, which is simply the equation expressing

the First Law of thermodynamics: the variation of the total energy of a system is accounted for

by all types of works upon the system and by the heat transfers. The level of readership of this

article is intended for undergraduate students and our aim is to help students and professors to

better address thermodynamical issues in the context of mechanical problems.

The classical dynamics of a system of particles of mass M is governed by Newton’s Second Law

which can be expressed by~Fext dt = Md~vcm , (1)

where ~Fext is the resultant of the external forces and ~vcm the center-of-mass velocity. The equation,

valid in a given inertial reference frame, remains invariant upon inertial frame transformations. The

previous equation also allows us to conclude that the infinitesimal variation of the kinetic energy

of the centre-of-mass equals the so called “pseudo-work” [4]. Actually, by multiplying both sides

of eq. (1) by ~vcm one concludes that the dot product of the resultant of the external forces by the

infinitesimal displacement of the centre-of-mass is

~Fext · d~rcm =1

2Mdv2cm . (2)

1guemezj@unican.es2tmanuel@teor.fis.uc.pt

1

The infinitesimal “pseudo-work” is the dot product of the resultant force applied to a system and

of the corresponding infinitesimal centre-of-mass displacement. It should not be confused with the

infinitesimal “work” of a force which is the dot product of the force and of its own infinitesimal

displacement. The above two equations are equivalent, though they may provide complementary

information. The first equation will be referred to as the impulse-momentum equation and states

that the infinitesimal impulse of ~Fext equals the infinitesimal variation of the linear momentum of

the center-of-mass, d~pcm = Md~vcm. The second equation, which will be referred to as the centre-

of-mass equation, states that the infinitesimal pseudo-work equals the infinitesimal variation of the

centre-of-mass kinetic energy, dKcm. Note that, in general, Kcm is not the total kinetic energy of

the system, K, which may also include internal kinetic energy, Kint, for instance, rotational energy.

Of course, from eqs. (1) and (2), one readily obtains the corresponding integral equations.

The resultant force acting on the system can be the sum of conservative and non-conservative

forces. On the other hand, if there are, for instance, thermal effects, these are not described by the

centre-of-mass equation (2) [5, 6]. Therefore, in order to study the motion of extensive deformable

bodies one needs additional equations to describe rotations [7] and processes involving production

[8] and dissipation [9] of mechanical energy. To keep the discussion in this paper at a simple level we

do not consider rotations. Here we are particularly interested in addressing the physical description

of the kind of processes that also need the First Law of thermodynamics [10].

For any system, its internal energy variation, dU , receives contributions from the variation of

the internal kinetic energy, dKint (including rotational kinetic energy or kinetic energy with respect

to the centre-of-mass), from any internal work, dwint = −dΦ (i.e. work performed by the internal

forces) [11], from the internal energy variations related to temperature variations, McdT (c is the

specific heat), from the internal energy variations related to chemical reactions [12], etc.

In thermodynamics both work and heat contribute to the internal energy variation of a system.

For a general process on a macroscopic deformable body, whose analysis needs to combine mechanics

and thermodynamics [6], instead of the centre-of mass equation one should consider the equation

dKcm + dU =∑

j

~Fext,j · d~rj + δQ (3)

which is nothing but an expression of the First Law of thermodynamics [13, 14]: It is more general

than the centre-of-mass equation (2), but both are valid. In (3) the infinitesimal heat is denoted

by a δ since it is not an exact differential. Each term in the sum on the right-hand side of eq. (3)

is work associated with each external force ~Fext,j = (Fx, j , Fy, j , Fz, j), and d~rj = (dxj,dyj,dzj) is

the infinitesimal displacement of the force ~Fext,j itself (and not, anymore, the displacement of the

centre-of-mass). Therefore, δWj = ~Fext,j · d~rj is actual work and not pseudo-work. The above

equations are valid in any inertial frame. This means that the amount of information provided by

the set of equations is exactly the same, irrespective of the chosen inertial frame — there are no

privileged inertial frames and the above equations are Galilean invariant.

In many situations there are forces with zero spatial displacement in a certain inertial reference

frame (e.g. the force exerted by the ground on the foot of a walking person). In these cases the force

produces no work but there is an impulse and a corresponding linear momentum variation: ~Fdt =

d~p. Let us consider an example from thermodynamics: a gas of small mass (hence, gravitational

forces are negligible) is in equilibrium inside a vertical cylinder enclosed by a piston at the top. A

force ~F = (0,−F, 0) is applied to the piston and, hence, by the piston to the gas. This force is

2

always compensated by an opposite force, −~F , exerted upon the gas by the bottom of the cylinder

(the horizontal forces produced by the cylinder walls have zero resultant and can be ignored).

The displacement of this force is always zero. For a quasi-static process the impulse-momentum

equation (1) and the centre-of-mass equation (2) applied to this system yields

{

Mdvcm = (F − F )dt1

2Mdv2cm = (F − F )dycm

, (4)

leading to dvcm = 0: if the cylinder is initially at rest, vcm = 0 during the process, though the

position of the centre-of-mass might change, dycm 6= 0.

On the other hand, in the energy equation (3), which incorporates the First Law of thermody-

namics, one has dKcm = 0 and, since the bottom force does not do any work, one obtains

dU = −Fdy + δQ (5)

(dy is the displacement of the force exerted by the piston) equivalent to the well known equation

dU = −PdV + δQ (here, P is the pressure and V is the volume).

For a body that behaves like an elementary particle (no rotation, no deformation, etc.) the

centre-of-mass equation and the First Law of thermodynamics provide the same information. But

when the body does not behave like an elementary particle, such as the gas in the cylinder, the two

equations provide complementary information.

Summarizing, besides the impulse-momentum equation, one has at our disposal the centre-of-

mass equation and the energy equation [15]. Both equations are always valid and, therefore, they

should be compatible. They may provide the same information if the problem under consideration

is a purely mechanical one. However, if the situation is within the scope of thermodynamics, the

second equation, which embodies the First Law, is more general and provides new information with

respect to the centre-of-mass equation. Of course, one still needs the Second Law of thermodynam-

ics, which states that, observed the First Law, only processes compatible with a non-decrease of the

entropy of the universe, ∆SU ≥ 0, are allowed. To illustrate the usage of the impulse-momentum

equation together with the centre-of-mass or with the more general energy equation, we discuss,

in the next sections, a set of examples ranging from mechanics to thermodynamics. We start

with a pure mechanical example, continue with problems involving destruction or production of

mechanical energy and conclude with a typical thermodynamical problem.

This paper is organized as follows. In section 2 we consider the motion of a block sliding down an

incline without friction. In section 3 we study a totally inelastic collision. In section 4 we study an

example of mechanical energy production by considering the motion of a person on rollers pushing

against a wall. In section 5 we treat an inherently inelastic process – the pulling of a chain. Finally,

in section 6 we address a typical thermodynamical problem – the Joule-Thomson experiment. With

these examples, ranging from a pure mechanical situation to a typical thermodynamical one, we

show the appropriateness of the First Law of thermodynamics with respect to the Newton’s Second

Law in the mechanical description of the system.

3

N

r

Gr

y

x

a

Figure 1: Rigid block on an incline.

2 Block on an incline

A rigid block of mass M slides down an incline without friction, as Fig. 1 shows. We describe

the process in a reference frame with horizontal x and vertical y. There are two forces, namely

the weight, ~G = (0,−Mg, 0) and the normal force, ~N = (N sin α,N cos α, 0), where α is the

angle between the incline and the horizontal plane. The block is a rigid body and there are no

rotations, hence the displacement of each force is also the displacement of the centre-of-mass. The

infinitesimal displacements are

d~rG = d~rN = d~rcm = (dL cos α,−dL sin α, 0) (6)

with dL the elementary displacement along the incline. In the absence of rotations or any other

change in the internal energy of the body (and no heat transfers) [16], the energy equation (3) is

exactly the centre-of-mass equation (2).

For the finite process, the block, initially at rest, slides down a distance L during the time

interval t0. The integration of the impulse momentum and of the centre-of-mass equations are

straightforward and lead to

N = Mg cosα ; t0 =

(

2L

g sin α

)1/2

; v0 = (2Lg sin α)1/2 . (7)

where v0 is the final centre-of-mass velocity. These results were obtained in the chosen reference

frame, but any other inertial frame could be used, leading to equivalent expressions. In particular

the frame with x along the incline and y perpendicular to it is commonly used.

The simple mechanical process studied in this section is reversible, i.e., if the body were launched

with opposite velocity from the basis of the incline, it would reach the same original height. This

means that ∆SU = 0. This is not the case if there were kinetic friction and it is also not the case

in some of the following examples.

3 Inelastic collision

The next example is a perfect inelastic collision of a plasticine ball with the ground [17] and the

situation is represented in Fig. 2. We are only interested in the part of the motion after the initial

contact with the soil, until the ball stops completely.

There are two forces: the weight, ~G, and the force, ~N , exerted by the ground, which, for

simplicity, is assume to be constant (can be regarded as an average force). With the y axis directed

upwards, the two forces are ~G = (0,−Mg, 0) and ~N = (0, N, 0) .

4

Gr

Nr

y

i f

y

h0vr

Figure 2: Inelastic impact of a plasticine ball with the floor (i – initial contact; f – ball stops).

We denote by h the vertical displacement of the centre-of-mass and by t0 the time interval

since the initial contact of the ball with the floor until its gets totally stopped. The displacements

associated with the forces and with the centre-of-mass are (note that the displacement of the normal

force is zero [7]):

∆~rG = ∆~rcm = (0,−h, 0) ∆~rN = (0, 0, 0). (8)

After integration, the centre-of-mass equation (2) leads to ∆Kcm =(

~G+ ~N)

· ∆~rcm and,

denoting by vcm,i = −v0 the velocity of the centre-of-mass when the ball reaches the floor, one

obtains (vcm,f = 0), for the impulse-momentum and center-of-mass equation,

Mv0 = (N −Mg)t0 ,

1

2Mv20 = (N −Mg)h .

(9)

Note that the second equation, namely 1

2Mv20 +Mgh = Nh, is not a ‘work-energy’ equation since

Nh is pseudo-work.

Let’s now exploit the energy equation (3), which can also be readily integrated (forces are

constant), assuming the form

∆Kcm +∆U = ~G ·∆~rG + ~N ·∆~rN +Q . (10)

In the absence of other effects (rotations, work of internal forces) the variation of the internal energy

of the ball is assumed to be

∆U = Mc(Tf − Ti) , (11)

where Ti,f are the initial and final temperatures of the ball and c its specific heat. This and Q

in (10) accounts for thermal effects in this type of inelastic collisions. The impulse-momentum

equation (1) and eq. (10) allow us to write

Mv0 = (N −Mg)t0 ,

−1

2Mv20 +Mc(Tf − Ti) = Mgh+Q .

(12)

To experimentally confirm these equations, now one would need a thermometer, in addition to a

watch and a ruler.

The two sets of equations (9) and (12) are both valid and must be compatible. From the

second equations of each set one concludes that Mc(Tf − Ti) = Q + N h [18]. Assuming that the

5

temperature change of the ball is negligible, Tf ≃ Ti ≃ T , one concludes that there is an energy

transfer, Q = −Nh, from the system. This energy is thermal energy (heat) exchanged with the

thermal reservoir at temperature T and, therefore, it is a positive quantity from the reservoir point

of view. The variation of the entropy of the universe is solely the variation of the entropy of the

thermal reservoir and it is given by

∆SU =Nh

T> 0 . (13)

For a deformed ball initially at rest on the ground, one could think about an injection of the amount

of energy N h from the heat reservoir, e.g. through the action of the force N . This process, opposite

to the one that we have studied, would be allowed by the First Law of thermodynamics. However,

it would lead to a decrease of the entropy of the universe, which is prohibited by the Second Law

of thermodynamics [19]. Therefore this inverse process cannot happen spontaneously [20]. This is

not the case for the pure mechanical example considered in the previous section, which does not

increase the entropy.

Let us now consider the description of the same inelastic process in the reference frame S′,

moving vertically, along the common y′y axes, with velocity V with respect to S.

The forces are identical in both reference frames but the displacements in S′, ∆~r ′, differ from

(8) in the y component, where there is an extra term −V t0:

∆~r ′

G = ∆~r ′

cm = (0,−h − V t0, 0) , ∆~r ′

N = (0,−V t0, 0) (14)

The impulse-momentum and the center-of-mass equations in the new frame are

Mv0 = (N −Mg)t0 ,

Mv0V + 1

2Mv20 = (N −Mg)(h + V t0) .

(15)

which reduces exactly to (9). Formally, in any situation, the equations in reference S′ are always

linear combinations of the equations in reference S, and vice-versa, according to the Principle of

Relativity, and we have explicitly shown this for a particular case. Therefore, by changing to a new

inertial reference frame, one does not gain or loose information (though one reference frame might

be better suited than another one regarding technical aspects)

Next we explore the energy equation also in S′:

∆K ′

cm +∆U ′ = W ′

G +W ′

N +Q′ , (16)

The internal energy variation and the heat are the same in both frames: ∆U ′ = ∆U and Q′ = Q.

The impulse-momentum and the energy equation are readily obtained and altogether lead to

−MV +M(v0 + V ) = (N −Mg)t0

−1

2Mv20 −MV v0 +Mc(Tf − Ti) = Mgh+MgV t0 −NV t0 +Q .

(17)

Again, these equations reduce to (12) as expected. The discussion on thermal effects, heat transfer

and increase in the entropy of the universe, presented above, still applies here — the temperature

and the entropy variation, ∆SU, are Galilean invariants. It is worth noticing that eq. (3) can still

be written as dKcm−∑

j~Fext,j · d~rj = dU − δQ and both sides of the equation are, per se, Galilean

invariant too.

6

4 Mechanical energy production

A boy on rollers, initially at rest, pushes against a wall (with infinite mass) and slides away over the

floor [8]. The mass of the system is M and no dissipative forces between the rollers and the floor are

considered. An external horizontal force, ~F , is exerted on him by the wall and, as a consequence,

his translational kinetic energy increases. The vertical forces — the weight and the normal force

from the ground — cancel each other and they play no role in the discussion. The force ~F does

not do any work because its displacement is zero [17] (Fig. 3).

x

0vr

D x c m

x

Fr

i f

Figure 3: Person on rollers pushing against a wall.

This is again a one dimensional problem and we describe it in reference frame S with the x axis

pointing to the right. The force along the x direction is ~F = (F, 0, 0), represented in the figure.

We assume F to be constant (F can be considered an average force). The displacements of the

centre-of-mass and of the horizontal force are

∆~rcm = (∆xcm, 0, 0) , ∆~rF = (0, 0, 0) (18)

where ∆xcm is the spatial displacement of the centre-of-mass occuring in the time interval t0, the

duration of the action of the horizontal force upon the person. Denoting by v0 the velocity of

the boy’s centre-of-mass when the contact with the wall ceases, the impulse-momentum and the

centre-of-mass equation lead to

Mv0 = Ft0 ,

1

2Mv20 = F∆xcm .

(19)

We stress that the second equation is not an expression of an ‘energy-work’ theorem since the

external normal force exerted by the wall does not produce any work [21]. If the force does not do

any work, where is the source of energy responsible for the increase of the centre-of-mass kinetic

energy?

These energetic issues are better discussed using the energy equation. During the process of

pushing against the wall, the work of ~F vanishes but not its impulse. There are biochemical

reactions in the person’s muscles. When a chemical reaction ξ is produced inside the body, these

may cause variations in its internal energy, ∆Uξ, volume, ∆Vξ, entropy, ∆Sξ, etc. If a chemical

reaction takes place at constant external pressure, P , in diathermal contact with a heat reservoir at

temperature T , part of the internal energy is used for the expansion against the external pressure

7

and part must be exchanged with the heat reservoir in order to ensure that the entropy of the

universe does not decrease. The work, WP , and the heat, Qξ, are:

WP = −P∆Vξ ; Qξ = T∆Sξ , (20)

When ∆Vξ < 0, the external pressure performs work on the system and when ∆Sξ > 0 the heat

reservoir increases the internal energy of the system. We assume that no other energy is exchanged

between the person and the environment (no extra heat is exchanged between the person and the

environment).

The energy equation for this process — eq. (3) — is:

∆Kcm +∆Uξ = WF +WP +Q , (21)

and, together with the impulse-momentum equation,

Mv0 = Ft0 ,

1

2Mv20 +∆Uξ = −P∆Vξ + T∆Sξ .

(22)

The second equation provides new information with respect to the second equation in (19). In

particular, the second equation in eq. (22) can also be expressed as

1

2Mv20 = −∆Gξ , (23)

where:

∆Gξ = ∆Uξ + P∆Vξ − T∆Sξ , (24)

is the Gibbs free energy variation, which is symmetric of the pseudo-work associated with ~F . This

equation shows that, in effect, the person increases its centre-of-mass kinetic energy thanks to the

internal biochemical reactions [12]. For given temperature and pressure, the variation of the Gibbs

free energy is the maximum useful work that can be obtained: Wmax = −∆Gξ.

In the reference frame S′, in standard configuration with respect to S, the centre-of-mass equa-

tion is

∆K ′

cm = ~F ·∆~r ′

cm , (25)

which, again with the impulse-momentum equation results in (in S′ the displacement of the force

exerted on the person’s hand is −V t0)

−MV −M(v0 − V ) = Ft0 ,

1

2MV 2 − 1

2M(v0 − V )2 = F (∆xcm − V t0) ,

(26)

equivalent to (19). The same happens with the energy equation in reference frame S′: from ∆K ′

cm+

∆U ′

ξ = W ′

F +W ′

P +Q′ , one obtains (22).

The process studied in this section implies no entropy increase of the universe and, therefore, it

is reversible. In fact, the kinetic energy of the system can be completely used, at least in principle,

to increase by −∆Gξ the free energy of any chemical reaction. Real bodies, which are articulated

or made of elastic materials, may acquire accelerations (changes of centre-of-mass momentum and

kinetic energy) as a consequence of the action of internal forces when they are in contact with an

infinite mass body (a wall, the floor, etc.); of course, internal forces, by themselves, cannot accelerate

8

the center-of-mass of an isolated body [22]. The final kinetic energy in the example studied in this

section, or in the case of a person that jumps, is not due to the work of the contact forces since they

have zero displacement [23]. Any explicit or implicit assumption that forces with zero displacement

applied to a moving body (forces on the foot of a person that walks, on the tires of a car that

accelerates, on the foot of a person that jumps, etc.) do work is incorrect. In fact, the total kinetic

energy of a composite, articulated body in contact with an infinite mass body can change, even

when no work is done by the external forces applied to it. In the description of processes with

production of mechanical energy, the role of the chemical reactions (food consumption in people

that walk or jump, fuel combustion in cars, etc.) plays a central role.

In this section we analised a case of mechanical energy production. In examples with mechanical

energy destruction (e.g. the plasticine ball that stops and deforms after its impact with the floor,

or the inherently elastic process of pulling a chain to be treated in the next section) we come out

to the conclusion that the entropy of the universe increases.

5 Pulling a chain

A chain of length L and mass density λ lies in a heap on the floor [24]. Its end is grabbed and pulled

horizontally with a force ~F , as Fig. 4 shows. As a result, every piece of the moving unwrapped

chain acquires a constant velocity ~v0.

L

L / 2

x

x

x

Fr

0vr

Figure 4: A chain unwraps under the action of a force ~F moving with constant velocity v.

We denote by x the position of the end of the chain, so that L − x is the length of the chain

that remains on the heap. The centre-of-mass of the chain is located at position (we assume that

the heap is narrow)

xcm =λx(x/2)

λL=

x2

2L. (27)

The centre-of-mass velocity is the time derivative and it is given by

vcm =xv0L

. (28)

The centre-of-mass momentum (product of the total mass by the centre-of-mass velocity) is

pcm = λLvcm = λx v0 = p . (29)

9

meaning that the linear momentum of the moving part of the chain is always equal to the linear

momentum of the centre-of-mass. The centre-of-mass kinetic energy is (half of the total mass times

v2cm)

Kcm =1

2λLv2cm =

1

2λx2v20L

, (30)

which is, in general, different from the total kinetic energy given by (half of the mass of the already

moving part times v20) :

K =1

2λxv20 . (31)

When the end of the chain moves by dx, the above mentioned magnitudes change accordingly,

and their infinitesimal changes are given by

dxcm =x

Ldx , (32)

dvcm =1

Lv0 dx , (33)

dpcm = λv0 dx = dp , (34)

dKcm = λx

Lv20 dx , (35)

dK =1

2λv20 dx . (36)

This is a one dimensional motion and the force is given by ~F = (F, 0, 0). Its infinitesimal displace-

ment and the infinitesimal centre-of-mass displacement, occurring in time dt are

d~rF = (dx, 0, 0) d~rcm =

(

x

Ldx, 0, 0

)

. (37)

The infinitesimal variation of the linear momentum and of the centre-of-mass kinetic energy [see

eqs. (34) and (35)] immediately lead to the following set of equations

λv0dx = Fdt

λ xLv

20dx = F x

L dx .(38)

Finally, one concludes that the pulling force must be constant and given by [25]

F = λv20 . (39)

The energy equation is dKcm + dU = ~F · d~rF + δQ. If one assumes that the chain does not

change its temperature during the process, the internal energy is then just the internal kinetic

energy, i.e. the kinetic energy relative to the centre-of-mass) and therefore dU = dK − dKcm

(Koning’s theorem [26]). Hence

dK = ~F · d~rF + δQ (40)

where dK is the kinetic energy variation. This equation, together with the impulse-momentum

equation [see (34) and (36) ] allows us to arrive at

λv0 dx = F dt

1

2λv20 dx = F dx+ δQ .

(41)

10

From the first equation we obtain the information already included in (38): with dx = v0dt, one

gets (39). From the second equation in (41) one obtains

1

2λv20 dx = λv20 dx+ δQ . (42)

Therefore, thermal effects are accounted for through the equation:

Q =

∫

δQ = −1

2λv20

∫ x

0

dx = −1

2λv20x = −

1

2Fx. (43)

This means that half of the work, Fx, produced by the pulling force is dissipated as thermal energy,

with the corresponding increase of the entropy of the universe.

The displacement of chains — articulated bodies formed by rigid links and with no internal

sources of energy —, is an inherently irreversible process. There are links of the chain that vary

their speeds abruptly, and these processes destroy mechanical energy [15]. Each link in the chain

that is, abruptly, placed into motion can be regarded as an inelastic collision, producing a loss of

mechanical energy and making the process irreversible [25].

The description of the process in reference S′ in standard configuration is also very instructive

for the illustration of the Principle of Relativity at work.

6 The Joule-Thomson experiment

William Thomson (who became Lord Kelvin) proposed a method to study the dependence of

the internal energy of a gas with its volume. Together with Joule, they carried out a series of

experiments using a device sketched in Fig. 5 [27]. A cylindrical tube, with cross-section of area

A, is divided into two parts by a porous-plug. In the Joule-Kelvin (or porous plug) process a gas

(usually a real gas), of mass M , occupying the initial volume Vi, on the left hand side, and pressure

Pi, is forced to pass through a porous plug or a valve against a final constant pressure Pf < Pi,

reaching a final volume Vf, on the right hand side. During the process the pressures on the left and

on the right (see figure) are kept constant but the gas temperature may change from Ti to Tf.

Fr

R

Fr

R

Fr

L

Fr

L

AP

i , V

i

Pf , V

f

p p

i

f

Figure 5: The Joule-Thomson or porous-plug (pp) experiment.

There are three forces acting on the gas: the left side force, ~FL = (APi, 0, 0), the right side

force, ~FR = (−APf, 0, 0), and the zero displacement force, ~N = (N, 0, 0), exerted on the gas by the

porous plug [28]. The displacements are

∆~rL = (Vi/A, 0, 0) , ∆~rR = (Vf/A, 0, 0) , ∆~rN = (0, 0, 0) , ∆~rcm = (∆xcm, 0, 0) . (44)

11

We denote by ∆xcm the horizontal displacement of the centre-of-mass, ∆xcm = (Vi + Vf)/(2A)

during the time interval t0, the duration of the process. For the sake of generality let us consider

initial and final centre-of-mass velocities denoted by vi,f (in a more restricted analysis they could

be simply set to zero). The impulse momentum equation, (~FL + ~FR + ~N)t0 = ∆~pcm and the

centre-of-mass equation, ∆Kcm = (~FL + ~FR + ~N) ·∆~rcm yield

M (vf − vi) = (PiA− PfA+N) t0

1

2M

(

v2f− v2

i

)

= (PiA− PfA+N)∆xcm .(45)

If the initial and final centre-of-mass velocities are zero, both equations lead to the same equation

for N , namely

N = (Pf − Pi)A (46)

i.e. the force exerted upon the gas by the porous-plug is directed to the left because Pf < Pi.

Now we consider the energy equation, ∆Kcm +∆U = ~FL ·∆~rL + ~FR ·∆~rR + ~N ·∆~rN +Q , and

assume zero centre-of-mass velocity [29], hence eq. (46) holds. In such case,

0 = (PiA− PfA+N) t0

∆U = PiVi − PfVf +Q .(47)

The Joule-Thomson experiment is carried on under adiabatic conditions, i.e. the device (cylin-

der and pistons) is thermally isolated so there is no heat involved in the process. In this case the

second eq. (47) reduces to

Uf − Ui = PiVi − PfVf (48)

and, since the enthalpy is related to the internal energy through H = U + PV , one concludes that

the process is isoenthalpic: Hf = Hi.

We can finally show that the entropy of the universe increases in this irreversible process. From

dH = V dP + TdS, for an isoenthalpic process

dSH = −V

TdPH > 0 (49)

since dPH is always negative (the pressure decreases in the process). The Joule-Thomson exper-

iment is a typical thermodynamical process in which all external forces have zero resultant (and

information resulting from the centre-of-mass equation).

If the process is studied in reference frame S′, moving with velocity V along the horizontal

direction, the porous-plug moves with velocity −V and the result set of equations is

0 = (PiA− PfA+N) t0

∆U = PiVi − PfVf − V (FL − FR +N) t0 +Q ,(50)

totally compatible with the equations in reference S.

12

7 Conclusions

The centre-of-mass equation, resulting from the integration of the Newton’s Second Law, and the

First Law of thermodynamics, establishing the energy balance in a process, correspond to two

absolutely different physical hypothesis. Both equations are valid but their applicability deserves

special attention. We addressed this issue, in a pedagogical way, by firstly presenting the equations

and then applying them to selected examples. Situations for which there is mechanical energy

production are particularly interesting. These include, among others, problems related with human

activities (a person jumping, climbing, cycling, etc.) or with living beings (a horse pushing a chariot,

etc.) and with thermal engines (a car that accelerates due to an internal combustion, etc.), which

are sometimes incompletely or even incorrectly addressed. This was actually a motivation for the

present work where we have discussed interesting physical situations ranging from mechanics to

thermodynamics.

Together with the impulse-momentum vector equation we used a scalar energy equation, which is

either the centre-of-mass equation or the First Law of thermodynamics, emphasizing the additional

information that can be obtained from the second one with respect to the other. We also paid some

attention to the frame transformation: the set of equations appropriate to discuss a problem provide

the same information irrespective of the chosen inertial reference frame (Principle of Relativity).

This was illustrated using various examples. Of course, the choice of the reference frame is usually

determined by the simplicity of the resulting formalism.

Intrinsic dissipative processes, such as processes with chains, should be related with thermo-

dynamics when they are solved. When a chain is unwrapped or raised, a significant part of the

work provided by the external force that displaces the chain is dissipated as heat, an effect that

deserves formalization into an equation, and which is not included when attention is only paid to

the mechanical part of the problem. We also considered a couple of examples involving forces acting

without displacing their application points. In a certain reference frame, such a zero displacement

force produces no work but it does have an impulse. In another inertial frame there might be both

impulse and work. Whenever pertinent we discussed other thermodynamical aspects, namely the

irreversibility of some processes in connection with the Second Law. On the contrary, we have

discussed the forces that act in thermodynamical processes, something that is really uncommon

in thermodynamical textbooks. For instance in the Joule-Kelvin process the porous plug exerts a

force on the gas and, even though it does not do any work, it plays a crucial role in the description

of the process.

Through the discussion of the five examples presented in this article we aimed at clarifying the

additional information provided by the more general First Law of thermodynamics with respect to

Newton’s Second Law.

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